题目链接

判断一颗二叉树是否是二叉搜索树(二叉排序树),也就是BST

如果该二叉树是BST, 那么对其中序遍历,所得序列一定是单调递增的(不考虑有重复数值的情况)

附上代码:

 /**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
// "fa" holds the last value that has been visited
// "flag" is false when it(the given binary tree or its subtree) is an invalid BST
void InOrder(TreeNode *root, int& fa, bool& flag) {
if (root->left != NULL) {
InOrder(root->left, fa, flag);
}
if (root->val <= fa)
flag = false;
fa = root->val;
if (root->right != NULL) {
InOrder(root->right, fa, flag);
}
}
bool isValidBST(TreeNode *root) {
if (root == NULL || root->left==NULL&&root->right==NULL) return true;
// initialize "fa" as INT_MIN
// I assume that there are no tree node's val equals to INT_MIN
// and it does... (test case like this doesnt exist)
int fa = INT_MIN;
bool flag = true;
InOrder(root, fa, flag);
if (flag)
return true;
else
return false;
}
};

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